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Let $R$ be a PID. Is every finitely generated projective $R[T]$-module free? In other words, is every vector bundle on $\mathbb{A}^1_R$ trivial?

For $R=k[X]$ this is true by the Theorem of Quillen-Suslin. If it fails in general, what happens for $k[X,X^{-1}]$?

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Just one variable (it should stay a PID). – Martin Brandenburg Feb 5 '13 at 19:26
Quillen-Suslin is for any (PID)$[x_1, \dots, x_n]$. – user18119 Feb 5 '13 at 21:13
@QiL: Oh, really? Do you have a reference for that? – Martin Brandenburg Feb 10 '13 at 14:33
See Quillen's Inventiones paper in 1976, Or Ferrand's Bourbaki talk. – user18119 Feb 11 '13 at 21:57
up vote 3 down vote accepted

Quote from Lam's Serre's problem on projective modules:

In 1958, Seshadri showed that Serre's conjecture is true for two variables (i.e. for $A = k[x_1,x_2]$). In fact, Seshadri proved that f.g. projectives over R[t] are free if $R$ is any commutative PID.

Reference: Seshadri, C.S., Triviality of vector bundles over the affine space $K^2$, Proc. Nat. Acad. Sci. USA 44, 456-458.

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As the answer is entirely included in that Seshadri's article, I realise I should delete any reference to Lam's book. But this book is so wonderful (as is anything written by Lam) I won't do so. – PseudoNeo Feb 5 '13 at 10:25
Great! Thank you. – Martin Brandenburg Feb 5 '13 at 11:22
Why is it enough to prove the case n=2 in Proposition 1? – Martin Brandenburg Feb 5 '13 at 19:26

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